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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>From the initial condition, we have</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\bf x}(0)=\left(
\begin{array}{c}
2\\
0\\
1
\end{array}
\right)=C_1  \left(
\begin{array}{c}
0\\
-2\\
1
\end{array}
\right) +C_2\left(
\begin{array}{c}
1\\
1\\
0
\end{array}
\right) +C_3 \left(
\begin{array}{c}
2\\
2\\
1
\end{array}
\right)  \to 
\begin{array}{c}
C_2+2C_3=2\\
-2 C_1+C_2+2C_3=0\\
C_1+C_3=1
\end{array}
\to C_1=1, C_2=2, C_3=0.
\end{equation*}
</div>
<p class="continuation">Thus, the solution for the initial value problem is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\bf x}=\left(
\begin{array}{c}
0\\
-2\\
1
\end{array}
\right) e^{t}+2\left(
\begin{array}{c}
1\\
1\\
0
\end{array}
\right) e^{2t}.
\end{equation*}
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